Solve Problem 1 if the sides $\mathit{x}\mathbf{=}\mathbf{0}$and $\mathit{x}\mathbf{=}\mathbf{1}$are insulated (see Problems 2.14 and 2.15), and $\mathit{T}\mathbf{=}\mathbf{0}$for $\mathit{y}\mathbf{=}\mathbf{2}$, $\mathit{T}\mathbf{=}\mathbf{1}\mathbf{-}\mathit{x}$ for$\mathit{y}\mathbf{=}\mathbf{0}$.

It has been given that the rectangular plate is covering the area $0<x<1$, $0<y<2$, if $T=0$ for $x=0$, $x=1$, $y=2$and $\text{T=1-x}$for $y=0$.

Laplace’s equation in cylindrical coordinates is,

$\begin{array}{c}{\mathbf{\nabla }}^{\mathbf{2}}\mathbf{u}\mathbf{=}\frac{\mathbf{1}}{\mathbf{r}}\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{r}}\mathbf{\left(}\mathbf{r}\frac{\mathbf{\partial }\mathbf{u}}{\mathbf{\partial }\mathbf{r}}\mathbf{\right)}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{r}}^{\mathbf{2}}}\frac{{\mathbf{\partial }}^{\mathbf{2}}\mathbf{u}}{\mathbf{\partial }{\mathbf{\theta }}^{\mathbf{2}}}\mathbf{+}\frac{{\mathbf{\partial }}^{\mathbf{2}}\mathbf{u}}{\mathbf{\partial }{\mathbf{z}}^{\mathbf{2}}}\\ \mathbf{=}\mathbf{0}\end{array}$

And to separate the variable the solution assumed is of the form$\mathit{u}\mathbf{=}\mathit{R}\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathit{\Theta }\mathbf{\left(}\mathbf{\theta }\mathbf{\right)}\mathit{Z}\mathbf{\left(}\mathbf{z}\mathbf{\right)}$.

Assume that the plate is so long compared to its width that the mathematical approximation can be made that it extends to infinity in the y direction, the temperature T satisfies Laplace's equation inside the plate that is${\nabla }^{2}T=0$.

To solve this equation, try a solution of the form mentioned below. $T(x,y)=X\left(x\right)Y\left(y\right)$ ….. (1 $Y\frac{d^{2}X}{d{x}^2}+X\frac{d^{2}Y}{d{y}^2}=0$ ….. (2)

Divide the above equation by XY

$\frac{1}{X}\frac{d^{2}X}{d{x}^2}+\frac{1}{Y}\frac{d^{2}Y}{d{y}^2}=0$ ….. (3)

Now, following the process of separation of variables, it can be written

$\begin{array}{c}\frac{1}{X}\frac{d^{2}X}{d{x}^2}=-\frac{1}{Y}\frac{d^{2}Y}{d{y}^2}\\ =\text{constant}\\ =-k^2,k\ge 0\end{array}$ ….. (4)

$X^{\text{'}\text{'}}=-k^{2}X$and$Y^{\text{'}\text{'}}=k^{2}Y$.

Here, the constant $k^2$is called the separation constant. The solutions for these equations for$k\ne 0$.

role="math" localid="1664343493282" $\begin{array}{l}X\left(x\right)=\left\{\begin{array}{c}\sin \left(kx\right)\\ \cos \left(kx\right)\end{array}\right\}\\ Y\left(y\right)=\left\{\begin{array}{c}\sinh \left(ky\right)\\ \cosh \left(ky\right)\end{array}\right\}\end{array}$ ….. (5)

The solution for $k=0$.

$\begin{array}{l}X\left(x\right)=\left\{\begin{array}{l}1\\ x\end{array}\right\}\\ Y\left(y\right)=\left\{\begin{array}{l}1\\ y\end{array}\right\}\end{array}$ ….. (6)

Write the relevant boundary conditions.

$\begin{array}{l}T(x,0)=1-x\\ T(x,2)=0\end{array}$

$\begin{array}{c}\frac{\partial T}{\partial x}(0,y)=\frac{\partial T}{\partial x}(1,y)\\ =0\end{array}$

The boundary condition $T(x,2)=0$implies that $Y\left(2\right)=0$. Thus, the possible solutions for $Y\left(x\right)$are mentioned below.

$Y\left(y\right)=\left\{\begin{array}{l}\begin{array}{ll}\sinh k(2-y),& \text{for}k\ne 0\end{array}\\ \begin{array}{ll}2-y,& \text{for}k=0\end{array}\end{array}\right.$

The boundary condition $\frac{\partial T}{\partial x}(0,y)=0$implies the condition given below.

${\frac{\partial X}{\partial x}|}_{x=0}=0$

Write the possible solutions for $X\left(x\right)$.

$X\left(x\right)=\left\{\begin{array}{l}\begin{array}{ll}\cos \left(kx\right),& \text{for}k\ne 0\end{array}\\ \begin{array}{ll}1,& \text{for}k=0\end{array}\end{array}\right.$

The boundary condition is,

$\frac{\partial T}{\partial x}(1,y)=0$

This implies that${\frac{\partial X}{\partial x}|}_{x=1}=0$

Impose this on eq. (4), see that constant solution is still allowed for the case of $k=0$. If $k\ne 0$, then eq. (5) yields that $k=n\pi .$Using the above results, the form of $T(x,y)$prior to applying the fourth boundary condition.

$T(x,y)=c_0(2-y)+\sum _{n=1}^{\infty }{c}_n\sinh \left(n\pi (2-y)\right)\cos \left(n\pi x\right)$

Apply the fourth boundary condition $T=1-x$for $Y=2$, $b_0=4c_0$, $b_n=c_n\sinh \left(2n\pi \right)$, for$n=1,2,3,\dots$.

We have the equation mentioned below.

$1-x=\frac{b_0}{2}+\sum _{n=1}^{\infty }{b}_n\sinh \left(2n\pi \right)\cos \left(n\pi x\right)$

Invert the Fourier series to determine the coefficients.

$\begin{array}{c}{b}_0=2{\int }_0^1(1-x)dx\\ =1\end{array}$

$\begin{array}{c}{\text{b}}_{\text{n}}=\frac{1}{2}{\int }_0^1(1-x)\cos \left(n\pi x\right)dx\\ =2\left(\frac{1-\cos \left(n\pi \right)}{{\left(n\pi \right)}^2}\right)\\ =\frac{1}{{\left(n\pi \right)}^2}(1-{(-1)}^n)\\ =\{\begin{array}{l}0\text{if}n\text{iseven}\\ \frac{2}{{\left(n\pi \right)}^2}\text{ifnisodd}\end{array}\end{array}$

Write final $T(x,y)$expression.

Inserting $b_0$and $b_n$into the above expression of $T(x,y)$it ends up with

$T=\frac{1}{4}(2-y)+\frac{4}{{\pi }^2}\sum _{\text{odd}n}\frac{1}{n^2\sinh \left(2n\pi \right)}\sinh n\pi (2-y)\cos \left(n\pi x\right)$

Hence, this is the steady-state temperature distribution.